a point charge of 5.0×10⁻⁷ c moves to the right at 2.6×10⁵ m/s in a magnetic field that is directed into the…

a point charge of 5.0×10⁻⁷ c moves to the right at 2.6×10⁵ m/s in a magnetic field that is directed into the screen and has a field strength of 1.8×10⁻² t. what is the magnitude of the magnetic force acting on the charge? 0 n 2.3×10⁻³ n 23 n 2.3×10¹¹ n

a point charge of 5.0×10⁻⁷ c moves to the right at 2.6×10⁵ m/s in a magnetic field that is directed into the screen and has a field strength of 1.8×10⁻² t. what is the magnitude of the magnetic force acting on the charge? 0 n 2.3×10⁻³ n 23 n 2.3×10¹¹ n

Answer

Explanation:

Step1: Recall the formula for magnetic force

The formula for the magnetic force on a moving - charge is $F = qvB\sin\theta$, where $q$ is the charge, $v$ is the velocity, $B$ is the magnetic - field strength, and $\theta$ is the angle between the velocity vector and the magnetic - field vector.

Step2: Determine the value of $\theta$

The charge moves to the right and the magnetic field is directed into the screen. The angle between the velocity vector and the magnetic - field vector $\theta = 90^{\circ}$, and $\sin\theta=\sin90^{\circ}=1$.

Step3: Substitute the given values into the formula

Given $q = 5.0\times10^{-7}\ C$, $v = 2.6\times10^{5}\ m/s$, and $B = 1.8\times10^{-2}\ T$. $F=qvB\sin\theta=(5.0\times 10^{-7}\ C)\times(2.6\times 10^{5}\ m/s)\times(1.8\times 10^{-2}\ T)\times1$. First, multiply the numerical values: $5.0\times2.6\times1.8 = 23.4$. Then, multiply the powers of 10: $10^{-7}\times10^{5}\times10^{-2}=10^{-7 + 5-2}=10^{-4}$. So, $F = 23.4\times10^{-4}\ N=2.34\times10^{-3}\ N\approx2.3\times10^{-3}\ N$.

Answer:

$2.3\times 10^{-3}\ N$